[tex]\text{Given that,}\\\\\cos \theta = -\dfrac 25 \\\\\implies \cos^2 \theta = \dfrac 4{25}\\\\\\\implies 1- \sin^2 \theta = \dfrac{4}{25}\\\\\\\implies \sin^2 \theta = 1-\dfrac{4}{25}\\\\\\\implies \sin^2 \theta =\dfrac{21}{25}\\\\\\\implies \sin \theta = \pm \sqrt{\dfrac {21}{25}} = \pm \dfrac{\sqrt{21}} 5\\\\\text{Since}~ \tan \theta>0,~ ~\text{the value of} ~\sin \theta ~\text{wii be:} \\\\\sin \theta = -\dfrac{\sqrt{21}}5[/tex]
